From 61056d57fa2c4ece7377d7736c07e8b0f12bb2af Mon Sep 17 00:00:00 2001
From: Prefetch
Date: Sun, 21 Feb 2021 19:28:33 +0100
Subject: Add "Legendre transform"

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 latex/know/concept/legendre-transform/source.md | 79 +++++++++++++++++++++++++
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+% Legendre transform
+
+
+# Legendre transform
+
+The **Legendre transform** of a function $f(x)$ is a new function $L(f')$,
+which depends only on the derivative $f'(x)$ of $f(x)$, and from which
+the original function $f(x)$ can be reconstructed. The point is, just
+like other transforms (e.g. Fourier), that $L(f')$ contains the same
+information as $f(x)$, just in a different form.
+
+Let us choose an arbitrary point $x_0 \in [a, b]$ in the domain of
+$f(x)$. Consider a line $y(x)$ tangent to $f(x)$ at $x = x_0$, which has
+a slope $f'(x_0)$ and intersects the $y$-axis at $-C$:
+
+$$\begin{aligned}
+    y(x) = f'(x_0) (x - x_0) + f(x_0) = f'(x_0) x - C
+\end{aligned}$$
+
+The Legendre transform $L(f')$ is defined such that $L(f'(x_0)) = C$ (or
+sometimes $-C$ instead) for all $x_0 \in [a, b]$, where $C$ is the
+constant corresponding to the tangent line at $x = x_0$. This yields:
+
+$$\begin{aligned}
+    L(f'(x)) = f'(x) \: x - f(x)
+\end{aligned}$$
+
+We want this function to depend only on the derivative $f'$, but
+currently $x$ still appears here as a variable. We fix that problem in
+the easiest possible way: by assuming that $f'(x)$ is invertible for all
+$x \in [a, b]$. If $x(f')$ is the inverse of $f'(x)$, then $L(f')$ is
+given by:
+
+$$\begin{aligned}
+    \boxed{
+        L(f') = f' \: x(f') - f(x(f'))
+    }
+\end{aligned}$$
+
+The only requirement for the existence of the Legendre transform is thus
+the invertibility of $f'(x)$ in the target interval $[a,b]$, which can
+only be true if $f(x)$ is either convex or concave, i.e. its derivative
+$f'(x)$ is monotonic.
+
+Crucially, the derivative of $L(f')$ with respect to $f'$ is simply
+$x(f')$. In other words, the roles of $f'$ and $x$ are switched by the
+transformation: the coordinate becomes the derivative and vice versa.
+This is demonstrated here:
+
+$$\begin{aligned}
+    \boxed{
+        \dv{L}{f'} = \dv{x}{f'} \: f' + x(f') - \dv{f}{x} \dv{x}{f'} = x(f')
+    }
+\end{aligned}$$
+
+Furthermore, Legendre transformation is an *involution*, meaning it is
+its own inverse. Let $g(L')$ be the Legendre transform of $L(f')$:
+
+$$\begin{aligned}
+    g(L') = L' \: f'(L') - L(f'(L'))
+    = x(f') \: f' - f' \: x(f') + f(x(f')) = f(x)
+\end{aligned}$$
+
+Moreover, the inverse of a (forward) transform always exists, because
+the Legendre transform of a convex function is itself convex. Convexity
+of $f(x)$ means that $f''(x) > 0$ for all $x \in [a, b]$, which yields
+the following proof:
+
+$$\begin{aligned}
+    L''(f')
+    = \dv{x(f')}{f'}
+    = \dv{x}{f'(x)}
+    = \frac{1}{f''(x)}
+    > 0
+\end{aligned}$$
+
+Legendre transformation is important in physics,
+since it connects Lagrangian and Hamiltonian mechanics to each other.
+It is also used to convert between thermodynamic potentials.
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